Non-parallel rays just means the angles of the mirrors and lens are changed a little (by that same half-degree). Alternately, your lenses aren't the standard 'perfect' lens but instead precision crafted to counteract the non-parallel rays due to the moon not being at infinite distance. (the portion of the lens further away from being directly 'under' the moon would have to have a slightly higher/lower index of refraction; or you could use water and some fancy material to adjust the shape dynamically as the moon moves across the sky? -or just nail the moon in place for the duration of the experiment >.> This is a what-if after all)

It's just arbitrary numbers. But from what little I've seen of lenses, long focal lengths aren't the problem apart from needing very high precision on the lens shape, short ones are limited by refractive index. Besides, at higher image sizes, you just need more lenses. -> by tuning each lens position to put its focus at the object, the size of your object is only limited by the minimum image size at focus.

RE coupling:The difference with this setup and the single big lens is that you're not coupling the moon's surface to the target one time, each lens system effectively creates a separate image of the moon onto the object. EDIT: and a separate image of the object onto the moon.

jseah wrote:Non-parallel rays just means the angles of the mirrors and lens are changed a little (by that same half-degree).

That's still not really how it works, if I understand the way you're describing that.

Sticking with a mirror1, you can have a surface (continuous or 'Fresnel') designed to take perfectly parallel rays cast out from a corresponding mirror-sized 'slice' of the Moon's surface, directly along the main axis of the mirror, and concentrate them to a 'point' (with or without additional intervening optics to make use of it outside of the axis, to least block the incoming image), or you can also design it to collect light rays from all the extreme outer limbs of the Moon via the (same, or opposite) outer edge of the mirror to take the whole Moon.

(Or, with a vastly larger-than-Moon mirror, also theoretically beyond the trivially visible limbs, but if it's 'looking at' more Moon surface than than the Sun is shining upon, it's not going to be significantly better.)

But the number of scattered rays that happen to conform to this exacting plan are limited. Most top-of-the-Moon light (that hits the mirror at all) is as likely to hit a 'non-equivalent' part of this idealised mirror and thus not redirect to the same spot as the 'proper' ray. This is how focusing actually works. The focuser is designed that (for a given distance of source object) all light emanating from a given point upon the source is focussed to an equivalent point upon the final focal plane, distinct from any other point on the focal plane that 'serves' all the captured incoming light from its own, different, source-point. (There's a complication that the outer-limb light from the Moon starts off roughly one Moon-radius further away than the centre-of-the-Moon does, of course, although the resulting blur is usually inconsequential in light of the total distance away, and could be fudged by a non-planar focal-surface, or something.)

Getting all of the captured 'parallel' (or tightly defined, angle-wise) light from all of the Moon onto point thus gives you all light that is not parallel(/as tightly defined) smudging off elsewhere. And getting all of the captured light from a given point-source mapped specifically onto a particular point-destination sends all different-point-source light to be mapped upon (at best, their own unique) different-point-destinations of focus.

Either way, all the light from all the visibly-shining Moon cannot find itself pumped into a single point, it's smeared out one way or another.

1 Easier to deal with mirrors, because chromatic aberrations through simple lenses can make things more complicated than a simple angle-of-incidence being opposite-and-equal to the angle-of-departure for all frequencies of light that reflect.

pileofrogs wrote:Hi all. I almost understood why the moon can't light a fire on the earth. I only have one question: Could I light a fire on earth with two moons?

More moons would help. They're still essentially poor reflectors, so a mere two still leaves you with a substantial engineering problem, and I can't think of reasonable orbits that would put two-similarly sized discs the size of our current moon in the sky conveniently close to each other, so we're still looking at rather more complex solutions than a mere magnifying glass.

Two moons shouldn't get you to hotter than the sun, though.

If you wanted to munchkin the "multiple mirrors pointed at different parts of the sun", a properly reflective dyson sphere should accomplish your goal, albeit with rather more significant engineering concerns.

Soupspoon wrote:Either way, all the light from all the visibly-shining Moon cannot find itself pumped into a single point, it's smeared out one way or another.

It isn't a single point in the plan I stated. The assumption is that the image can be focused to some small but non-zero size (1cm was given as an arbitrary number), which is then the size of your target.

Given that you can focus the sun down to a spot of roughly that size, I'd assume you can do the same to the moon's image too.

ijuin wrote:You can focus the sun to a spot of that size with a small lens, but not necessarily with a lens dozens of meters across.

Two same-size lenses with different focal lengths can focus the sun to the same spot size at different distances. Which would allow you to overlap the spots. (see diagram)

This diagram? With the partially-refracting merging lens? That also defeats the perfect reversibility, as it would be if it's a partially-reflecting mirror (which would be perpendicular to that lens, to achieve the same)...

jseah wrote:The lines past the mirror was drawn in to show the path without the mirror. It's supposed to reflect off the mirror there to the target.

If so, I misunderstood (but still don't see the point of) that diagram. You were trying to get multiple images 'overlaid', yet that mirror is only there some of the time, so there's only one image..?

EDIT: was reversibility a requirement of the system? I wouldn't think a multi lens setup is reversible either since you don't get a coherent image on the target.

Reversibility (definite ray from a definite angle from the definite source to a definite angle at the definite destination could as easily be at a definite angle from definite destination to definite source again) is the point. As soon as you start trying to combine rays, in one direction, you have to be randomly distributing a ray in multiple other directions, in the other, which isn't a 'free action', which is the conceit of a major part of the article...

why isn't this a "free action"? isn't that basically what solar collector farms do? they direct the light from one source (sun) via multiple mirrors to one spot. in a completely passive setup, ignoring tracking.

As I said a while back, I call BS on the thermodynamics argument that it is impossible per se to concentrate the energy of moonlight to generate a temperature hotter than the surface of the moon as it is clearly possible to collect the energy of moonlight (using, e.g., photovoltaics) and concentrate them at a point (e.g., using a laser powered by the photovoltaics) to create a hotter temperature in a small area, albeit by cooling a much larger area.

There may be valid optics arguments as to why this is not possible using mirrors/lenses, and these may rely on optics enforcing thermodynamics, but there is not a thermodynamics argument blocking it via every possible route since doing it in some ways does not violate thermodynamics. Unfortunately Randall's explanation of this was pretty unclear which is why this subject has led to so much scepticism.

Soupspoon wrote:If so, I misunderstood (but still don't see the point of) that diagram. You were trying to get multiple images 'overlaid', yet that mirror is only there some of the time, so there's only one image..?

Soupspoon wrote:Reversibility (definite ray from a definite angle from the definite source to a definite angle at the definite destination could as easily be at a definite angle from definite destination to definite source again) is the point. As soon as you start trying to combine rays, in one direction, you have to be randomly distributing a ray in multiple other directions, in the other, which isn't a 'free action', which is the conceit of a major part of the article...

If you place multiple beams of light such that they all focus onto a small volume (but as noted, they spread out on the other side of that volume, since they cannot be travelling in the same direction, and indeed cannot all be from the same focal length)and then place a black body in that volume that absorbs the incoming light, that body can end up absorbing far more rays.

You also get the same point on the body receiving rays from multiple directions. (ie. multiple images)

FOARP wrote:As I said a while back, I call BS on the thermodynamics argument that it is impossible per se to concentrate the energy of moonlight to generate a temperature hotter than the surface of the moon as it is clearly possible to collect the energy of moonlight (using, e.g., photovoltaics) and concentrate them at a point (e.g., using a laser powered by the photovoltaics) to create a hotter temperature in a small area, albeit by cooling a much larger area.

There may be valid optics arguments as to why this is not possible using mirrors/lenses, and these may rely on optics enforcing thermodynamics, but there is not a thermodynamics argument blocking it via every possible route since doing it in some ways does not violate thermodynamics. Unfortunately Randall's explanation of this was pretty unclear which is why this subject has led to so much scepticism.

It's not possible using a single imaging solution.

Which is interesting and educational, but definitely not the only possible solution.

FOARP wrote:As I said a while back, I call BS on the thermodynamics argument that it is impossible per se to concentrate the energy of moonlight to generate a temperature hotter than the surface of the moon as it is clearly possible to collect the energy of moonlight (using, e.g., photovoltaics) and concentrate them at a point (e.g., using a laser powered by the photovoltaics) to create a hotter temperature in a small area, albeit by cooling a much larger area.

There may be valid optics arguments as to why this is not possible using mirrors/lenses, and these may rely on optics enforcing thermodynamics, but there is not a thermodynamics argument blocking it via every possible route since doing it in some ways does not violate thermodynamics. Unfortunately Randall's explanation of this was pretty unclear which is why this subject has led to so much scepticism.

Check footnote [2]:"And, more specifically, everything they do is fully reversible—which means you can add them in without increasing the entropy of the system."

This is the difference between optics and photovoltaics, and why the thermodynamic argument doesn't immediately fail.

Soupspoon wrote:Either way, all the light from all the visibly-shining Moon cannot find itself pumped into a single point, it's smeared out one way or another.

It isn't a single point in the plan I stated. The assumption is that the image can be focused to some small but non-zero size (1cm was given as an arbitrary number), which is then the size of your target.

Given that you can focus the sun down to a spot of roughly that size, I'd assume you can do the same to the moon's image too.

However you manage it, the fact remains that all you can do with lenses and mirrors is expand the apparent size of the light source.

Filling an entire sphere with the moon gives you a bit more than twice the light of the Sun, for an equilibrium temperature of about 1.2 times what something can have at noon.

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You can't use lenses and mirrors to make something hotter than the surface of the light source itself.

If you're surrounded by the bright surface of the Moon, what temperature will you reach? Well, rocks on the Moon's surface are nearly surrounded by the surface of the Moon, and they reach the temperature of the surface of the Moon (since they are the surface of the Moon.) So a lens system focusing moonlight can't really make something hotter than a well-placed rock sitting on the Moon's surface.

The Moon isn't a light source, it's a bad mirror. Moonlight is reflected Sunlight. The temperature of Buzz Aldrin on the Moon is dependent on how long he sits in the Sunlight. So is the temperature of the surrounding rocks. This is reflected in the design of the Lunar Suits. Place a small off axis parabolic reflector of sufficient size on the surface of the moon with a lens at the focus and I would have thought you could raise the temperature of a piece of paper to the ignition point with light from the lens. That would be moonlight would it not? Light reflected from the moon.

morriswalters wrote:My problem with the What If lies with these statements.

You can't use lenses and mirrors to make something hotter than the surface of the light source itself.

If you're surrounded by the bright surface of the Moon, what temperature will you reach? Well, rocks on the Moon's surface are nearly surrounded by the surface of the Moon, and they reach the temperature of the surface of the Moon (since they are the surface of the Moon.) So a lens system focusing moonlight can't really make something hotter than a well-placed rock sitting on the Moon's surface.

The Moon isn't a light source, it's a bad mirror. Moonlight is reflected Sunlight. The temperature of Buzz Aldrin on the Moon is dependent on how long he sits in the Sunlight. So is the temperature of the surrounding rocks. This is reflected in the design of the Lunar Suits. Place a small off axis parabolic reflector of sufficient size on the surface of the moon with a lens at the focus and I would have thought you could raise the temperature of a piece of paper to the ignition point with light from the lens. That would be moonlight would it not? Light reflected from the moon.

What am I missing?

The first quote is indeed a mistake, because it's true of emitters but not necessarily of reflectors. However, the second is still a valid observation even if not for exactly the reasons implied in the what-if.

If something facing a hemisphere of reflective lunar surface (e.g. something a bit above the lunar surface) can reach a particular temperature from the combination of the Sun and the lunar surface, then something on Earth exposed to lenses that expanded the moon to fill the entire sky (i.e. a hemisphere) would not get any hotter, even if the Sun's light (unmagnified) were added to the mix.

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I was wrong. It turns out that the Moon indeed is an emitter, I just hadn't thought it through. It emits in infrared, since it is at thermal equilibrium(more or less). This defines the limit, I think. 304W/M2(average).

tagno25 wrote:It is not the temperature of the moon that matters, it is the photon density. If it was just the temperature, then we could burn paper with starlight (or generate electricity).

It turns out that The Naked Scientists at Cambridge state that someone wrote a paper on how to light paper with a telescope doing exactly that.

morriswalters wrote:I was wrong. It turns out that the Moon indeed is an emitter, I just hadn't thought it through. It emits in infrared, since it is at thermal equilibrium(more or less). This defines the limit, I think. 304W/M2(average).

tagno25 wrote:It is not the temperature of the moon that matters, it is the photon density. If it was just the temperature, then we could burn paper with starlight (or generate electricity).

It turns out that The Naked Scientists at Cambridge state that someone wrote a paper on how to light paper with a telescope doing exactly that.

As the energy of the sun at the earth is ~1340 w/m2, that means we should be looking at 1340 w/m2 of energy coming off the moon (whether black body or reflected doesn't matter for this argument).

Therefore, at a close distance from the moon it is not actually important whether you're using sunlight or moonlight: either will be sufficient to start a fire (as sunlight is sufficient to start a fire).

Therefore the only question is whether the increased distance from the moon is enough to diffuse the moonlight such that it is no longer possible: but a lens of sufficient size and complexity would appear to solve that question.

morriswalters wrote:I was wrong. It turns out that the Moon indeed is an emitter, I just hadn't thought it through. It emits in infrared, since it is at thermal equilibrium(more or less). This defines the limit, I think. 304W/M2(average).

tagno25 wrote:It is not the temperature of the moon that matters, it is the photon density. If it was just the temperature, then we could burn paper with starlight (or generate electricity).

It turns out that The Naked Scientists at Cambridge state that someone wrote a paper on how to light paper with a telescope doing exactly that.

As the energy of the sun at the earth is ~1340 w/m2, that means we should be looking at 1340 w/m2 of energy coming off the moon (whether black body or reflected doesn't matter for this argument).

Therefore, at a close distance from the moon it is not actually important whether you're using sunlight or moonlight: either will be sufficient to start a fire (as sunlight is sufficient to start a fire).

Therefore the only question is whether the increased distance from the moon is enough to diffuse the moonlight such that it is no longer possible: but a lens of sufficient size and complexity would appear to solve that question.

I am again convinced that Randall is wrong on this one.

Let us know how you get on trying to light a fire using earthlight (which, by your argument, should be sufficient to start a fire at a close distance from the Earth). There's a big difference between 1340W/m2 of highly organised, very nearly parallel sunlight and 1340W/m2 of disorganised, diffuse earthlight...

Randall may be wrong, but so are you. No arrangement of lenses and reflectors can collect all the energy that comes off the Moon and direct it at a single point. You can either capture the light that comes from a smaller area in more directions, or you can capture the light that comes from a larger area in fewer directions, but you can't do both.

In addition to Earthlight as rmsgrey mentions (i.e. light that's reflected up from the ground to an object that is shaded from direct sunlight), by your logic we would also be able to start a fire from a totally black Moon. But the surface of a blackbody this distance from the Sun would only get up to 392K if it's emitting the full 1340W/m^2 (And as explained, you can't use optics to get something hotter than the emitter). The Moon reflects about 12% of incident light and doesn't conduct heat very quickly, which is why it's only about 100C under direct sunlight. (If it did conduct heat very quickly, then it would radiate the same amount from its night side as from its day side, and would be barely above freezing throughout.)

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rmsgrey wrote:Let us know how you get on trying to light a fire using earthlight (which, by your argument, should be sufficient to start a fire at a close distance from the Earth). There's a big difference between 1340W/m2 of highly organised, very nearly parallel sunlight and 1340W/m2 of disorganised, diffuse earthlight...

Well we're a very different case. The actual incident radiation at the earth surface (ground level) is well below 1340, and the earth conducts heat better and also spins (hence why earth soil temperatures max out a lot lower than the 100 degrees Randall mentioned for lunar rocks), meaning it doesn't get as hot.

But yeah, for me the optics arguments are much the stronger than the temperature argument.

The Solar Isolation number is taken as a projection on the surface of a plane. The moon is a sphere. Divide by 4. 340 is the right number. Cambridge University agrees with Randell, they suggest that the hottest that the paper could get is -50(I'm assuming degrees Celsius).

dawolf wrote:Well we're a very different case. The actual incident radiation at the earth surface is far below 1340, and the earth conducts heat better and also spins (hence why earth soil temperatures max out a lot lower than the 100 degrees Randall mentioned for lunar rocks).

No, the Moon spins as well. The number at the top of the atmosphere is the number to use. For all intents and purposes for heat energy the atmosphere is the fuzzy planetary boundary. And the Earth must radiate away that solar energy no less than the Moon. And that is precisely the problem with carbon dioxide. It changes the equilibrium temperature.

It's about the photon. It's about having more, hotter photons coming in that going out.

Dr What and jseah were on the mark in previous posts. The "color" (frequency) of a photon is equivalent to its energy. The light coming from the moon is infrared because the moon is cold. If not then the new moon would glow. ( it does reflect some earthlight but thats another topic. ) The moonlight that hits us is partly from the moon but mostly white hot sunlight reflected off of the moon surface. Those photons will impart kinetic energy into what they hit equall to the kinetic energy they absorbed from the source, i.e., the 'temperature' will be the same. Period. If you want to say that somehow the moon is able to reduce the energy of a photon without changing its frequency then watch out because you're about to get hit in the head with a big fat quantum mechanics book. Dimmer light is not cooler light. Redder light is cooler light.

It all comes down to how many hot photons come in and how many go out. The photon has enough 'temperature', but you need a lot of them to make enough heat. We don't catch fire when standing in sunlight even though the photons are 'hot' enough to do so because we radiate and conduct the heat away quickly. Mostly its heat conduction because of contact with air. If no air then its harder to cool ( thats why the shuttle had to open its doors to expose radiators to cool itself in the vacuum of space. http://spaceflightnow.com/shuttle/sts107/030225amos/ir.html.)

Once your temperature is as hot as the light source hitting you, you're going to radiate heat as fast as you absorb it. That's why a pot in a kiln become invisible once they reach the same temp as the kiln, everything is glowing the same color of red, because its all in equilibrium. If the pot glows brighter than the kiln, it will heat the kiln, and things balance out again. This is the thermodynamic argument. You can't make your paper hotter than the sun.

So the trick is to get enough photons coming in to out number the ones going out and overcome the energy lost via conduction. A magnifying glass on paper works because with sunlight, you get a lot of hot photons.

Yes, focusing light down to an arbitrary point can't be done with optics. But no, that doesn't mean I can't get a whole lot of photons onto a piece of paper. It doesn't matter that some or even most of them miss the paper because I can't focus. I can get 500 square miles of postage stamp sized mirrors to reflect moonlight that all converges on a nearly postage sized stamp ( they'll all spread a little, and some will come in a oblique angles, ) but I promise you that spot is going to get a whole lot of white hot photons that will torch your piece of paper in no time. In the end, it comes down to what percentage of a sphere surrounding the target is used to direct light, leaving the unused part of that sphere 'open' to space to radiate heat away. If the paper is fully surrounded ( imagine using fiber optics ) then it will heat to the temperature of the sun.

If I hung the paper in a little glass test tube full of air, and then put that test tube in a glass chamber with no air, I could eliminate most of the conduction heat losses and, in principle, do the same with star light. Yep. I went there.

the limiting factor is the temperature of the light which is the temperature of the source. Optics and reflections are only practical ( but significant) difficulties.

It's about the photon. It's about having more, hotter photons coming in that going out.

Dr What and jseah were on the mark in previous posts. The "color" (frequency) of a photon is equivalent to its energy. The light coming from the moon is infrared because the moon is cold. If not then the new moon would glow. ( it does reflect some earthlight but thats another topic. ) The moonlight that hits us is partly from the moon but mostly white hot sunlight reflected off of the moon surface.

Yes, a photon has an inherent energy, but the integration of the number of photons and the wavelength is what matters. A blue (high energy) photon has the same energy in full sunlight as it does in moonlight but there are a lot more of them in sunlight. The moon's cross section receives the energy per unit area that the earth's cross section does. However, the moon reflects that light as a sphere. That means that the sun's parallel rays are spread out as they reflect back towards earth, and only a small fraction of them actually encounter the earth, net result, fewer photons per unit area.

In theory, if a flat section of the moon was a perfect specular reflector, and directly overhead during the full moon, you would get all of the sun's energy reflected. However, the moon is more of a diffuse reflector than a specular one, and an imperfect one at that. That means that any light that hits the surface tends to come out as uniform spherical radiation regardless of the position of the sun. (This is why the edges of the full moon look about as bright as the center).

Yes, full sunlight at local noon on the moon's equator is just as intense as full sunlight at local noon on the earth's (and with no atmosphere to distribute heat, cooling in space, or on the moon, is a major challenge). But full moonlight at local midnight directly under the moon is not.

And no, it is not possible to sufficiently concentrate that light back down to a tight enough focus to reach the temperature of the original source. Optical limitations prevent that. It doesn't matter how well you insulate your paper, or how perfect an absorber it is, it will radiate off sufficient energy in the infrared that it will reach equilibrium below ignition temperature, despite the individually higher energy of the incoming photons.

Yeah, as mentioned repeatedly, the best you can do with reflectors is make a larger area just as bright as your source. If you fill the sky with small, appropriately angled mirrors, then you end up with a piece of paper facing a hemisphere of moonlight. Which is the same situation as something a bit above the surface of the Moon that is shaded from the Sun. You won't get any hotter than that using optics.

If it's a full sphere, you could get 19% hotter, but that'll only ignite paper if the surface of the moon is about 400K to start with, which I'm pretty certain is not the case.

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gmalivuk wrote:If it's a full sphere, you could get 19% hotter, but that'll only ignite paper if the surface of the moon is about 400K to start with, which I'm pretty certain is not the case.

Sources do say the moon is about 400K. For kicks I tried to get a reading with my pyrometer but not surprisingly I couldn't get a could reading much less trust it. Regardless, I'm curious, where do you get 19% from?

Power is proportional to the fourth power of temperature, so if the power doubles (which is what happens if you go from a hemisphere to a full sphere), then the temperature goes up to 2^(1/4) what it was, which is about 1.19.

Space.com says the temperature of the Moon gets up to 396K. Multiplied by 2^.25 gives 471K, or 198ºC. Wikipedia says the auto-ignition temperature of paper starts at 218ºC.

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gmalivuk wrote:Yeah, as mentioned repeatedly, the best you can do with reflectors is make a larger area just as bright as your source. If you fill the sky with small, appropriately angled mirrors, then you end up with a piece of paper facing a hemisphere of moonlight. Which is the same situation as something a bit above the surface of the Moon that is shaded from the Sun. You won't get any hotter than that using optics.

If it's a full sphere, you could get 19% hotter, but that'll only ignite paper if the surface of the moon is about 400K to start with, which I'm pretty certain is not the case.

There's a divergence here. The surface temperature of the moon is one thing, but there is also the reflected visible light that is getting dropped between the two paragraphs.

With an albedo of 12% for the moon, you're trying to light the paper with sunlight, after passing it through frosted welding glass, but at least the spectrum of incoming light is appropriate to the task.

Is adding filters and quantum mechanics to your mirrors and lenses allowed?

If you had one of those ideal thermodynamically isolated boxes with a window and a filter that reflects any light longer than UV, you'd get it up to temperature over an excessively long period of time with that sunlight reflected off the moon. That's a lot more expensive than dollar store magnifying glasses, unfortunately.

The surface of the moon is probably not a significantly different temperature than a shaded object slightly above the surface of the moon, which would account for all the reflected sunlight.

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gmalivuk wrote:The surface of the moon is probably not a significantly different temperature than a shaded object slightly above the surface of the moon, which would account for all the reflected sunlight.

Some posters clearly disagree with that assumption.

morriswalters wrote: It turns out that the Moon indeed is an emitter, I just hadn't thought it through. It emits in infrared, since it is at thermal equilibrium(more or less). This defines the limit, I think. 304W/M2(average).

With 12% albedo vs 1360 W/m^2, that would give ~160W/m^2 for the reflected light.

If those numbers are correct, the solar reflection is a significant portion (1/3rd) of the power coming off the moon.

Still classically useless in distribution, but not negligible as such.Also, it will not be a blackbody spectrum, so the effective temperature won't be the same as a blackbody with the same wattage.

gmalivuk wrote:The surface of the moon is probably not a significantly different temperature than a shaded object slightly above the surface of the moon, which would account for all the reflected sunlight.

Some posters clearly disagree with that assumption.

Sure, and some people think the Earth is flat.

I have not seen a convincing argument against the assumption, though, while the what-if itself has a convincing argument for it (i.e. rocks sticking up from the surface are exposed to roughly a hemisphere of Moon-reflected sunlight, therefore their temperature already takes that reflection into account).

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